What's the best explanation of the fallacy in this 'paradox'?

Question

Of course whenever you have two statements that each on its own sound plausible but then contradict each other, you can simply check which one is false by e.g. drawing a picture. But I hope that someone can find a clear explanation that addresses the heart of the misconception. So here is the paradox.

We know (for $x \to +\infty)$) that

$e^{-x}$ goes to zero faster than the reciprocal of any polynomial. $\tag{1}$

In particular we have that $-e^{-x}$ (which happens to be the derivative of $e^{-x}$) increases towards 0 faster than $-\frac{1}{(x+1)^2}$ (which happens to be the derivative of $\frac{1}{x + 1}$).

This means that the graph of $e^{-x}$ becomes flatter and flatter faster than that of $\frac{1}{x + 1}$ and since both start at $(0, 1)$ and tend to zero we find that $e^{-x} \geq \frac{1}{x + 1}$ and hence in particular:

$\frac{1}{x + 1}$ goes to zero faster than $e^{-x}$, $\tag{2}$

contradicting (1). What gives?

Answer 1

You got me for a minute until I plotted these on Desmos.

As you can see, $-\frac{1}{(x+1)^2}$ initially increases faster than $-e^{-x}$. Then, it was overtaken at around $x=2.513$.

This is enough to explain your "paradox", because $e^{-x}$ initially decreases faster than $\frac{1}{x+1}$.

In general, asymptotics only apply "for large enough $x$" and therefore should not generate paradoxes like yours.

Answer 2

I have re-written without the Paradox :

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Claim X : Curve $C1 :: e^{-x}$ goes to zero faster than the reciprocal of any polynomial. [ true ]

In particular we have that $-e^{-x}$ (which happens to be the derivative of $e^{-x}$) increases towards 0 faster than $-\frac{1}{(x+1)^2}$ (which happens to be the derivative of $\frac{1}{x + 1}$). [ true ]

This means that the graph of $e^{-x}$ becomes flatter and flatter faster than that of Curve $C2 :: \frac{1}{x + 1}$ ... [ true ]

... and since both start at $(0, 1)$ and tend to zero we find that $e^{-x} \geq \frac{1}{x + 1}$ ... [ not true ]
... and since both start at $(0, 1)$ and tend to zero we find that $e^{-x} \leq \frac{1}{x + 1}$ ... [ true ]

... and hence in particular:

... $\frac{1}{x + 1}$ goes to zero faster than $e^{-x}$ [ not true ]
... $e^{-x}$ goes to zero faster than $\frac{1}{x + 1}$ [ true ]

... contradicting Claim X. [ not true ]
... which is consistent with Claim X. [ true ]

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Elaboration :

When a curve becomes flatter , that means it is "reaching" the "Destination" limit.
When a curve becomes flatter faster , that means it is "reaching" the "Destination" limit faster.
When Curve $C1$ is "reaching" the "Destination" faster than Curve $C2$ , then Curve $C1$ must be closer to the Destination than Curve $C2$.

Destination in this Case is $0$ , hence [[ $e^{-x}$ is closer to $0$ ]] more than [[ $\frac{1}{x+1}$ is closer to $0$ ]] !

Putting that in Math Syntax , we will get necessary Conclusion without Paradox !

Answer 3

This viewpoint is at least implicit in the existing comments and answers, but maybe can be boiled down further:

Suppose Car 1 and Car 2 start at point $A$ at time $0$ and end at point $B$ at $t = 1$. Eventually Car 1 is slower than Car 2. Which car is ahead at the end of the race (while Car 1 is slower)?

If that doesn't clarify the intuition, run time backward from $t = 1$: Two cars start at $B$ at the same time. Which car is initially behind (i.e., closer to $B$), the slower Car 1, or the faster Car 2?

(Technical note: Using $t = 1$ versus letting $t \to \infty$ is immaterial. The closed unit interval $[0, 1]$ and the extended positive reals $[0, \infty]$ are equivalent as ordered sets.)